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Suppose we have $f: \mathbb{D} \rightarrow \mathbb{D}$ analytic and not identically zero. In order to prove $f$ has an infinite Blaschke product representation (where of course the product defines an analytic function on the disk), I need to show that $f$ necessarily has countably many zeros $a_j$ in $\mathbb{D}$, and that they satisfy $$\sum_{j=1}^{\infty} (1-|a_j|)<\infty.$$ This condition seems necessary because the Blaschke product formed by the zeros of $f$ converges locally uniformly on the disk if and only if they satisfy the summation condition, which is how one would show the product is analytic.

Assuming this crucial step, I am able to prove that $f$ has a representation $f(z)=B(z)g(z)$ where $B$ is a product as above and $g$ is a never vanishing analytic function. Any help is appreciated.

Rick Sanchez
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  • If there are uncountably many zeros, then (since $\mathbb{D}$ is $\sigma$-compact) the zeros have an accumulation point, hence $f$ is identically zero. – carmichael561 Apr 21 '16 at 04:22
  • An accumulation point in the open disk, not the boundary? – Rick Sanchez Apr 21 '16 at 04:24
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    Yes, in the disk. We can write $\mathbb{D}=\cup_{n=1}^{\infty}{|z|\leq 1-\frac{1}{n}}$, and if the set of zeros is uncountable then one of the disks ${|z|\leq 1-\frac{1}{n}}$ contains infinitely many of the zeros. – carmichael561 Apr 21 '16 at 04:27
  • This is a very good point, and this doesn't even require boundedness of $f$. I believe the summation condition does though – Rick Sanchez Apr 21 '16 at 04:28
  • @carmichael561 : no, $\mathbb{D} = { |z| < 1}$, consider $\sin(1/(z-1))$, a function analytic on a bounded open set $U$ can have a (countably) infinite number of zeros on $U$, in that case it has an essential singularity on the boundary (otherwise it is identically zero) . – reuns Apr 21 '16 at 04:38
  • (sorry I thought you said countably many zeros) following the usual argument of isolated zeros and that $\int_C \frac{f'(z)}{f(z)} dz$ is finite on any compact where $f(z)$ is analytic (and has no zero on the contour $C$), we get that an analytic function always has a finite number of zeros on any compact set, and hence (considering as carmichael561 said the compacts ${ |z| \le 1-1/n}$) that $f(z)$ has at most a countably infinite number of zeros on $\mathbb{D}$ (more generally on any open set where it is analytic) – reuns Apr 21 '16 at 04:43

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